Problem 4 - Advanced Engineering Mathematics

Question

In Problems, determine whether the given matrix \(\mathbf{A}\) is diagonalizable. If so, find the matrix \(\mathbf{P}\) that diagonalizes \(\mathbf{A}\) and the diagonal matrix \(\mathbf{D}\) such that \(\mathbf{D}=\mathbf{P}^{-1} \mathbf{A} \mathbf{P}\). $$ \left(\begin{array}{ll} 0 & 5 \\ 1 & 0 \end{array}\right) $$

Step-by-Step Solution

Answer

\( \mathbf{A} \) is diagonalizable with \( \mathbf{P} = \begin{bmatrix} 5 & 5 \\ \sqrt{5} & -\sqrt{5} \end{bmatrix} \) and \( \mathbf{D} = \begin{bmatrix} \sqrt{5} & 0 \\ 0 & -\sqrt{5} \end{bmatrix} \).

Step 1: Find Eigenvalues

First, determine the eigenvalues of matrix \( \mathbf{A} \). For a matrix \( \mathbf{A} \), eigenvalues can be found by solving the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). For matrix \( \begin{bmatrix} 0 & 5 \ 1 & 0 \end{bmatrix} \), this becomes \( \det\begin{bmatrix} -\lambda & 5 \ 1 & -\lambda \end{bmatrix} \), which simplifies to \( \lambda^2 - 5 = 0 \). Solving this gives \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \).

Step 2: Find Eigenvectors

For each eigenvalue, find the corresponding eigenvector. For \( \lambda = \sqrt{5} \), solve \((\mathbf{A} - \sqrt{5} \mathbf{I})\mathbf{v} = \mathbf{0}\). This gives the system of equations: \(-\sqrt{5} v_1 + 5 v_2 = 0\) and \(v_1 - \sqrt{5} v_2 = 0\). Solving these yields the eigenvector \( \mathbf{v}_1 = \begin{bmatrix} 5 \ \sqrt{5} \end{bmatrix} \). For \( \lambda = -\sqrt{5} \), solve \((\mathbf{A} + \sqrt{5} \mathbf{I})\mathbf{v} = \mathbf{0}\), giving the eigenvector \( \mathbf{v}_2 = \begin{bmatrix} 5 \ -\sqrt{5} \end{bmatrix} \).

Step 3: Form Matrix P

Place the eigenvectors found in Step 2 as columns in the matrix \( \mathbf{P} \). Thus, \( \mathbf{P} = \begin{bmatrix} 5 & 5 \ \sqrt{5} & -\sqrt{5} \end{bmatrix} \).

Step 4: Check Diagonalizability and Form Matrix D

Since matrix \( \mathbf{A} \) has 2 distinct eigenvalues and a corresponding set of 2 independent eigenvectors, \( \mathbf{A} \) is diagonalizable. Form \( \mathbf{D} \) with the eigenvalues along its diagonal: \( \mathbf{D} = \begin{bmatrix} \sqrt{5} & 0 \ 0 & -\sqrt{5} \end{bmatrix} \).

Step 5: Verify the Condition \( \mathbf{D} = \mathbf{P}^{-1} \mathbf{A} \mathbf{P} \)

Calculate \( \mathbf{P}^{-1} \) and multiply with \( \mathbf{A} \mathbf{P} \). Verify that the resulting matrix equals \( \mathbf{D} \) calculated earlier.

Key Concepts

Eigenvalues

Eigenvalues are a fundamental concept in linear algebra, specifically when dealing with matrices. An eigenvalue of a matrix \( \mathbf{A} \) is a scalar \( \lambda \) that satisfies the equation \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \), where \( \mathbf{v} \) is a non-zero vector called an eigenvector. To find the eigenvalues, we solve the characteristic equation \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \). The solution to this equation gives us the eigenvalues for the matrix.
For example, in the matrix \( \begin{bmatrix} 0 & 5 \ 1 & 0 \end{bmatrix} \), we solve \( \lambda^2 - 5 = 0 \), which provides the eigenvalues \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \). These eigenvalues are essential for determining the matrix's diagonalizability.

Eigenvectors

Once eigenvalues are determined, the next step is to find the corresponding eigenvectors. An eigenvector of a matrix \( \mathbf{A} \) corresponding to an eigenvalue \( \lambda \) is a vector \( \mathbf{v} \) such that \( \mathbf{A} \mathbf{v} = \lambda \mathbf{v} \). In practice, we solve \((\mathbf{A} - \lambda \mathbf{I}) \mathbf{v} = \mathbf{0} \) to find these eigenvectors.

For the given matrix with eigenvalues \( \sqrt{5} \) and \( -\sqrt{5} \), we solve two systems of equations. The solution gives us \( \mathbf{v}_1 = \begin{bmatrix} 5 \ \sqrt{5} \end{bmatrix} \) for \( \lambda = \sqrt{5} \) and \( \mathbf{v}_2 = \begin{bmatrix} 5 \ -\sqrt{5} \end{bmatrix} \) for \( \lambda = -\sqrt{5} \). These vectors are vital since they enable us to construct the matrix \( \mathbf{P} \) for diagonalization.

Diagonal Matrix

A diagonal matrix is a type of matrix where all the entries outside the main diagonal are zero. In essence, it is simple to work with since matrix operations like addition, multiplication, and finding determinants become straightforward. When diagonalizing a matrix \( \mathbf{A} \), we aim to represent it in the form \( \mathbf{P}^{-1} \mathbf{A} \mathbf{P} = \mathbf{D} \), where \( \mathbf{D} \) is the diagonal matrix.

In our example, the diagonal matrix \( \mathbf{D} \) is formed with the eigenvalues on the diagonal: \( \begin{bmatrix} \sqrt{5} & 0 \ 0 & -\sqrt{5} \end{bmatrix} \). Transforming a matrix into diagonal form is helpful because it simplifies many computations, such as raising the matrix to a power, solving linear systems, and more.

Characteristic Equation

The characteristic equation is an essential polynomial equation when finding eigenvalues of a matrix. It is defined as \( \det(\mathbf{A} - \lambda \mathbf{I}) = 0 \), where \( \mathbf{I} \) is the identity matrix, and \( \lambda \) represents an eigenvalue. Solving this equation provides us the eigenvalues necessary for diagonalization.

In the example of matrix \( \begin{bmatrix} 0 & 5 \ 1 & 0 \end{bmatrix} \), the characteristic equation becomes \( \det\begin{bmatrix} -\lambda & 5 \ 1 & -\lambda \end{bmatrix} = \lambda^2 - 5 = 0 \). By solving this polynomial, we identify the eigenvalues \( \lambda = \sqrt{5} \) and \( \lambda = -\sqrt{5} \). Understanding the characteristic equation is critical since it underpins the calculation of eigenvalues, which are integral to many areas in linear algebra.